Wavio Sequence
64-bit integer IO format: %lld Java class name: Main
Wavio is a sequence of integers. It has some interesting properties.
Wavio is of odd length i.e. L = 2*n + 1.
The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.
The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.
No two adjacent integers are same in a Wavio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to find out the length of the longest Wavio sequence which is a subsequence of the given sequence. Consider, the given sequence as :
1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1.
Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the output will be 9.
Input
The input file contains less than 75 test cases. The description of each test case is given below: Input is terminated by end of file.
Each set starts with a postive integer, N(1<=N<=10000). In next few lines there will be N integers.
Output
For each set of input print the length of longest wavio sequence in a line.
Sample Input
10
1 2 3 4 5 4 3 2 1 10
19
1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1
5
1 2 3 4 5
Sample Output
9
9
1
解题：最长上升子序列加强版。单调队列优化！！！重点。先顺着求最长上升子序列，再逆着求最长上升子序列。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 10100; 18 int dp1[maxn],dp2[maxn],d[maxn],q[maxn]; 19 int bsearch(int lt,int rt,int val) { 20 while(lt <= rt) { 21 int mid = (lt+rt)>>1; 22 if(q[mid] < val) lt = mid+1;//严格上升单调取少于符号，上升的取少于等于 23 else rt = mid-1; 24 } 25 return lt; 26 } 27 int main() { 28 int n,i,j,head,tail; 29 while(~scanf("%d",&n)) { 30 for(i = 0; i < n; i++) 31 scanf("%d",d+i); 32 head = tail = 0; 33 for(i = 0; i < n; i++) { 34 if(head == tail) { 35 q[head++] = d[i]; 36 dp1[i] = head-tail; 37 }else if(d[i] > q[head-1]){ 38 q[head++] = d[i]; 39 dp1[i] = head-tail; 40 }else{ 41 int it = bsearch(tail,head-1,d[i]); 42 dp1[i] = it - tail + 1; 43 q[it] = d[i]; 44 } 45 } 46 head = tail = 0; 47 for(i = n-1; i >= 0; i--) { 48 if(head == tail) { 49 q[head++] = d[i]; 50 dp2[i] = head-tail; 51 }else if(d[i] > q[head-1]){ 52 q[head++] = d[i]; 53 dp2[i] = head-tail; 54 }else{ 55 int it = bsearch(tail,head-1,d[i]); 56 dp2[i] = it - tail + 1; 57 q[it] = d[i]; 58 } 59 } 60 int ans = 1; 61 for(i = 0; i < n; i++){ 62 ans = max(ans,min(dp1[i],dp2[i])*2-1); 63 } 64 printf("%d ",ans); 65 } 66 return 0; 67 }